# Proving question

• March 28th 2009, 08:37 PM
noob mathematician
Proving question
Since E[Y] = E[E[Y/X]], can i infer that Y = E[Y/X]? Therefore i conclude that:

Cov(X, E[Y/X])
= E[(X - E[X])(E[Y/X] - E[E[Y/X]])]
= E[(X - E[X])(Y- E[Y])]
= Cov(X, Y)

• March 29th 2009, 09:47 AM
Laurent
Quote:

Originally Posted by noob mathematician
Since E[Y] = E[E[Y/X]], can i infer that Y = E[Y/X]?

No you can't. But, it is true that $E[f(X)E[Y|X]]=E[f(X)Y]$ for any function $f$ (that's a characterization of the conditional expectation).

Quote:

Therefore i conclude that:

Cov(X, E[Y/X])
= E[(X - E[X])(E[Y/X] - E[E[Y/X]])]
= E[(X - E[X])(Y- E[Y])]
= Cov(X, Y)

You can use the above equality with $f(X)=X-E[X]$ : it gives $E[(X-E[X])E[Y|X]]=E[(X-E[X])Y]$. Since additionaly $E[E[Y|X]]=E[Y]$, this justifies your second equality.